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Sauron [17]
3 years ago
9

What is the relationship between the frequency and the pitch of a sound?

Physics
1 answer:
jarptica [38.1K]3 years ago
6 0
The pitch of a sound come from the frequency of the soundwave, soundwaves that have bigger frequencies sound with a high pitch and soundwaves with less frequencie have low pitch.
You might be interested in
Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

d_{12} = \sqrt{3^{2}+3^{2}  }  = \sqrt{18} m : distance from q₁ to q₂

(d₁₂)² = 18 m²

d_{13} =\sqrt{1^{2}+3^{2}  } = \sqrt{10} m  : distance from q₁ to q₃

(d₁₃)² = 10 m²

d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

(d₁₄)² = 13 m²

K=  8.98755 × 10⁹ N *m²/C²

q₁=  7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

Fn₁= 23.95*10⁻⁹ N

8 0
3 years ago
A 562 N trunk is on frictionless plane inclined at 30.0 degrees from the horizontal. What is the acceleration of the trunk down
Len [333]

Answer: 0m/s²

Explanation:

Since the forces acting along the plane are frictional force(Ff) and moving force(Fm), we will take the sum of the forces along the plane

According newton's law of motion

Summation of forces along the plane = mass × acceleration

Frictional force is always acting upwards the plane since the body will always tends to slide downwards on an inclined plane and the moving acts down the plane

Ff = nR where

n is coefficient of friction = tan(theta)

R is normal reaction = Wcos(theta)

Fm = Wsin(theta)

Substituting in the formula of newton's first law we have;

Fm-Ff = ma

Wsin(theta) - nR = ma

Wsin(theta) - n(Wcos(theta)) = ma... 1

Given

W = 562N, theta = 30°, n = tan30°, m = 56.2kg

Substituting in eqn 1,

562sin30° - tan30°(562cos30°) = 56.2a

281 - 281 = 56.2a

0 = 56.2a

a = 0m/s²

This shows that the trunk is not accelerating

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3 years ago
Select the correct answer<br>What is the importance of writing a draft?​
givi [52]

Answer:to revise or edit

anything that can be made in the non draft one

Explanation:

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3 years ago
(10p+15,15-10q)=(25,5)​
Harrizon [31]
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If a coworker repeatedly asks you out for a date even though she knows you are not interested, what is she guilty of?
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D sexual harassment
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